The thing you say "can't be right" is wrong because you are failing to take into account the fact that if a ball is removed at stage 1, it can't also be removed at stage 2. So really it should be 1/10+9/10*1/19+181/190*1/28+...You are right that it's easier to compute the chance of a ball not being selected, but you have a sum where you should have a product:[math]p'=\prod \frac{9k}{9k+1}=\prod 1-\frac{1}{9k+1}[/math]The series [imath]\sum -\frac{1}{9k+1}[/imath] is comparable to the harmonic series, and so diverges to [imath]-\infty[/imath]. An elementary fact about infinite products (easily proven by taking logs) is that if [imath]\sum a_n \to \pm \infty[/imath], and |an|<1 for all n, then [imath]\prod (1+a_n) \to \infty[/imath] (if the sum diverges to positive infinity) or 0 (if the sum diverges to negative infinity). Thus the chance of any given ball never being removed is 0. A countable union of measure 0 events has measure 0, so at midnight, the jug will almost surely be empty.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

For c:There are no balls in the jug.Each ball has a chance of being removed, and so, over an infinite amount of interactions, each ball is removed.

For b:The jug has infinitely many balls in it.As soon as we put the numbers 1 to 10 in, we remove number 10. When numbers 11 to 20 go in, number 20 is removed. We are effectivley only putting in 9 balls each time.

For a:I think the jug is empty.I'm not sure about this, but it seems that, unlike B, over an infinite number of interactions, every ball will be removed.

[Edit]This reminds me of a similar puzzle I read a few years ago, where two people were slaving away in hell for all eternity. Every day, Satan would stop by and deliver their pay for that day. Alice would get 9 gold coins added to her pile of cash, while Bob would get 10 gold coins added, but Satan would randomly remove one coin (after adding the 10)On a day to day count, Alice and Bob were both recieving the same amount of money, but at the end of eternity (ha ha) Alice would be rich while Bob would have nothing.

skeptical scientist wrote:Thus the chance of any given ball never being removed is 0. A countable union of measure 0 events has measure 0, so at midnight, the jug will almost surely be empty.

You mean full.

No. Each ball has a probability of 0 of never being removed, so the event "there exists a ball which is never removed" has probability 0. Therefore the event "all balls are eventually removed" has probability 1, and the jug is empty.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

skeptical scientist wrote:Thus the chance of any given ball never being removed is 0. A countable union of measure 0 events has measure 0, so at midnight, the jug will almost surely be empty.

You mean full.

No. Each ball has a probability of 0 of never being removed, so the event "there exists a ball which is never removed" has probability 0. Therefore the event "all balls are eventually removed" has probability 1, and the jug is empty.

bogglesteinsky wrote:For c:There are no balls in the jug.Each ball has a chance of being removed, and so, over an infinite amount of interactions, each ball is removed.

But the chance of the ball being removed is not the same each time, so (while not the case here) there are probability distributions where there would be balls left at the end. Compare this to the fact that a random walk in 3 dimensions is not recurrent. There's always a chance you get back home, but even with infinite many tries, the chance of you making it is not 1.

For a:I think the jug is empty.I'm not sure about this, but it seems that, unlike B, over an infinite number of interactions, every ball will be removed.

Consider ball n. Can you figure out if/when ball n is removed?

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

bogglesteinsky wrote:For c:There are no balls in the jug.Each ball has a chance of being removed, and so, over an infinite amount of interactions, each ball is removed.

But the chance of the ball being removed is not the same each time, so (while not the case here) there are probability distributions where there would be balls left at the end. Compare this to the fact that a random walk in 3 dimensions is not recurrent. There's always a chance you get back home, but even with infinite many tries, the chance of you making it is not 1.

I see. It's obviously more complicated than I first thought.

mike-l wrote:

For a:I think the jug is empty.I'm not sure about this, but it seems that, unlike B, over an infinite number of interactions, every ball will be removed.

Consider ball n. Can you figure out if/when ball n is removed?

Yes, surely the nth ball is removed at the end of the nth interaction, so i guess that means the jug ends up empty

The thing you say "can't be right" is wrong because you are failing to take into account the fact that if a ball is removed at stage 1, it can't also be removed at stage 2. So really it should be 1/10+9/10*1/19+181/190*1/28+...You are right that it's easier to compute the chance of a ball not being selected, but you have a sum where you should have a product:[math]p'=\prod \frac{9k}{9k+1}=\prod 1-\frac{1}{9k+1}[/math]The series [imath]\sum -\frac{1}{9k+1}[/imath] is comparable to the harmonic series, and so diverges to [imath]-\infty[/imath]. An elementary fact about infinite products (easily proven by taking logs) is that if [imath]\sum a_n \to \pm \infty[/imath], and |an|<1 for all n, then [imath]\prod (1+a_n) \to \infty[/imath] (if the sum diverges to positive infinity) or 0 (if the sum diverges to negative infinity). Thus the chance of any given ball never being removed is 0. A countable union of measure 0 events has measure 0, so at midnight, the jug will almost surely be empty.

Just thought about this:

Couldn't this be like one of those "0 chance but not impossible" things? I mean, there is an infinity of balls we are computing the chance for. That means, that, even if the chance for each individual ball is 100%, because of the infinite number of balls we're considering, it might be that some balls might get away.

Couldn't this be like one of those "0 chance but not impossible" things? I mean, there is an infinity of balls we are computing the chance for. That means, that, even if the chance for each individual ball is 100%, because of the infinite number of balls we're considering, it might be that some balls might get away.

Its definitely one of those situations, but anything like this (with an infinite number of events) will be one of those situations.

Couldn't this be like one of those "0 chance but not impossible" things? I mean, there is an infinity of balls we are computing the chance for. That means, that, even if the chance for each individual ball is 100%, because of the infinite number of balls we're considering, it might be that some balls might get away.

Exactly, someone who couldn't see the numbers wouldn't be able to say how many balls were left, even if they had watched you putting them in and taking them out. The result shouldn't change under relabelling!

Indigo is a lie.Which idiot decided that websites can't go within 4cm of the edge of the screen?There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

It doesn't change under relabeling. It changes if the relabeled balls are removed in a different fashion after the relabeling. Either way the number of balls left is the same as the number put in and not later removed (obviously).

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

Macbi wrote:Exactly, someone who couldn't see the numbers wouldn't be able to say how many balls were left, even if they had watched you putting them in and taking them out. The result shouldn't change under relabelling!

If a person who couldn't see the numbers could know how many balls there were in the urn, then there would be a contradiction. The fact that they don't know how many are left when you deprive them of the labeling makes sense: with less information, less is known.

In some ways its an [imath]\infty - \infty[/imath] type situation. That expression is undefined for a reason.

Caesar wrote:I understand the solution (I even got it myself) but it's still hurting my brain. This seems like a problem you can't approach with a limit but with set theory. Is there a guideline to easily recognize what kind of infinity we're talking about in a given problem? If you know what I mean?

Its infinite sets, not infinity as a number, in this case, as you recognise. I don't have a nice rule for you, sorry.

If we look at the ratio of balls left in to balls taken out, every interaction, we get (10 balls in - the ball out)/(the ball out).For n interactions, this is 9n/n.Obviously, resolving this to n=infinity, we get 9(infinity/infinity), which cancels to 9(1/1), or 9.So even after an infinite number of interactions, we still have a ratio of nine balls in to 1 taken out.

It just doesnt seem right to me to resolve this type of question with a "lul but i is subtracting infinity from infinity wich is 0 so there be nun" equivalent.Lets say that we take out the lowest number ball at each interaction. Interaction 1, removed ball = 1, highest ball = 10. Interaction 10, removed ball = 10, highest ball = 100. For any arbitrarily large number of interactions, the number of balls remaining = 10n-n = 9n, lowest ball remaining: n+1, highest ball remaining = 10n. Suddenly claiming that there are 'no' balls remaining after an 'infinite' number of interactions may be mathematically possible, but since mathematics tends to have problems with the concept of infinity (or atleast in the methods you're using), not necessarily logically so.

As for what numbers are on the balls removed, it seems like an illogical question for a jar theoretically containing 'infinity' balls.

Of course, I'm obviously wrong with all of this and am going to be crucified because I didn't say 'probability', or show any understanding of statistics.Sorry.

skeptical scientist wrote:Blacksail, do you agree that the balls left at the end are exactly those that are put in at some stage and not later removed?

I understand the reasoning, that every ball must be taken out because otherwise another interaction would take out that ball, but I also dont see how that gets around my objection that at every step you add balls and the sum of positive numbers is positive.

Its a nonphysical situation though; obviously you cant put balls in or out at an arbitrary rate. So there, my answer is "do it and get back to me"

I hope you don't view this as crucifying you, Furanku, but I want to walk through the ideas that you've brought up.

Furanku wrote:If we look at the ratio of balls left in to balls taken out, every interaction, we get (10 balls in - the ball out)/(the ball out).For n interactions, this is 9n/n.Obviously, resolving this to n=infinity, we get 9(infinity/infinity), which cancels to 9(1/1), or 9.So even after an infinite number of interactions, we still have a ratio of nine balls in to 1 taken out.

I think this really clearly explains the way that everybody starts to think about the problem, and it makes a certain amount of sense. It gives the correct answer for problem b, for instance. But it gives the wrong answer when applied to problem a, which is something that I want to convince you of, and I want to explore where the error in the reasoning is. That the reasoning provides the right answer for b makes it more plausible, but on close examination flaws become apparent.

One flaw that appears on the surface is that [imath]\frac{\infty}{\infty}[/imath], like [imath]\infty-\infty,[/imath] is an indeterminate form, that is, it could be equal to just about anything. The way to resolve indeterminate forms is to get your hands dirty and work out what is actually going on.

In this situation, the way that I, and I assume others, formalise the situation is to work out what balls are in the jug after the nth interaction. As you point out, there are only a finite number of balls in the urn at that time, so I hope you would agree that the question makes sense. If we use the rule for ball removal given in problem a, and let An = "the numbers on the balls in the urn after the nth interaction" then

Now, if a ball is still in the urn after every step, then its in the urn in the the end. But, if there is some step such that after that step the ball is never in the urn, then, the ball can't be in the urn at the end. For instance, ball number 1 cannot be in the urn in the end because its not in the urn after the first step and nor is it in the urn at the end of any other step. If there are balls in the urn at the end, they must have numbers on them, and those numbers must be in every An except a finite number at the start. But, if you look at the sets An, you realise that every number is in only a finite number of those sets, so no ball is in the urn in the end.

Now, this is only a thought experiment. There are not urns like this and its impossible to do the things with the balls that the problem suggests. But I do think that the question "if there is a ball in the urn, what number is on the ball?" is a logical question to ask. You say it doesn't make sense because there are theoretically an infinite number of balls in the urn, but if I put all the prime numbered balls in the urn it would still make sense to ask what qualities a number on a ball in the urn has, despite there being an infinite number of balls in the urn.

Vicious Chicken wrote:A slight variation to case A:

Each step, you add nine balls (not ten), and instead of removing the lowest numbered ball, you get out a pen and add a '0' to the end of its number. So you add 1-9, and the 1 becomes 10; second step add 11-19, and the 2 becomes 20. So, at every step, the state of the jug (the numbers on the balls it contains) will be identical to case A, but since you never remove any balls, it seems even more nonsensical to claim the jug is empty at midnight. And yet...

I would actually argue that there are an infinite number of balls in the urn at the end of this process, but the "numbers" on them don't have a finite number of digits: they'll have an arbitrary string of digits followed by an infinite string of 0's.

Let's turn this 169-ish problem into something that actually makes mathematical sense. Let p(i,n) = probability of ball i being in the jug after n rounds. The "correct" solutions to this problem are really solutions to [math]\sum_{i=1}^{\infty}\lim_{n\rightarrow\infty}p(i,n)[/math] while some people instead interpreted it as [math]\lim_{n\rightarrow\infty}\sum_{i=1}^{\infty}p(i,n)[/math] What makes one interpretation better than the other?

Thank you for summaries portions of your argument, Jesting. It definately helps in thinking of these concepts.Still, I'm feeling stubborn, so I'll foolishly and blindly continue

jestingrabbit wrote:One flaw that appears on the surface is that [imath]\frac{\infty}{\infty}[/imath], like [imath]\infty-\infty,[/imath] is an indeterminate form, that is, it could be equal to just about anything.

Oh, definately. This was more to point out that the ratio of balls in to balls out, per iteration, remains constant. I doubt that many people here would argue against a number divided by itself equalling 1, even for arbitrarily large numbers. 2/2 = 1, 10/10 = 1, googol/googol = 1. It may have been a large mistake on my part to use [imath]\infty[/imath] as a substitute for rather large numbers, but I intended it to illustrate that, in each instant, the net rate of balls entering the jar remains constant.

I like your argument here:

jestingrabbit wrote:Now, if a ball is still in the urn after every step, then its in the urn in the the end. But, if there is some step such that after that step the ball is never in the urn, then, the ball can't be in the urn at the end.

Obviously at iteration n, we remove ball n, and thus remove the possibility of it being drawn out at midnight. This continues for every iteration, as you've stated. I have difficulty with your interpretation when n approaches infinity, though. Im sure you'll happily correct me when you come back to this, and Im sure that I've misunderstood you or others. To me, thought, it appears that at each iteration n (in question a), the ball n+1 remains in the jar, as do every other ball above it up to 10n. For any arbitrarily large number, this is true (Iteration Googolplex: the googolplexth ball is taken other, but ball 'One googolplex and one' remains, as do all balls between it and 'Ten googolplexes' inclusive), unless I've made an error which you will undoubtedly highlight for me. Now, it looks as if, from your point of view when we cross the mythical 'finite to infinite' iteration boundary, we suddenly get an admittedly silly '[imath]\infty-\infty,[/imath] = 0 balls left' scenario even though, at every finite point, we had a finite number of balls remaining in the jar. This finite number was on the rise too. As soon as you make n = infinity though, finite numbers go out the window, and asking for finite values becomes a bit pointless. The only way we can deal with this is by operating with 'infinities'. Im not even going to try to understand alephs or anything, but it seems to me there is going to be an infinite number of balls remaining in the jar, even if you cannot define the number on any ball picked from the jar after midnight. Whatever undefined quantity of balls that is removed at midnight, 9 times that undefined quantity remains. While trying to multiply infinity by anything is patently silly, so is trying to use remove infinity balls from an infinitly large jar filled with an infinite number of balls.

Dealing with infinity is always a bit ridiculous. And I can't remember what I was saying.

Goplat wrote:Let's turn this 169-ish problem into something that actually makes mathematical sense. Let p(i,n) = probability of ball i being in the jug after n rounds. The "correct" solutions to this problem are really solutions to [math]\sum_{i=1}^{\infty}\lim_{n\rightarrow\infty}p(i,n)[/math] while some people instead interpreted it as [math]\lim_{n\rightarrow\infty}\sum_{i=1}^{\infty}p(i,n)[/math] What makes one interpretation better than the other?

Firstly, I challenge any suggestion that this is a 169. Just because the conclusions are unexpected and initially appear paradoxical, doesn't mean that its just some trick of the language. By that standard, even Blue Eyes is a 169.

Secondly, we shouldn't worry about the probabilistic problem c. Blacksails and Furanku have issues with the solution that bogglesteinsky presented for problem a, and there are no probabilities there, so why don't we discuss the disagreement in the simplest situation in which the disagreement occurs. I would just point out, though, that just because there is a calculation that can be done, doesn't mean that the calculation has meaning (the case of the missing dollar has a good example of a calculation that has no meaning).

If every ball has a number on it, and you claim that there are an infinite number of balls in the urn, and what we are doing is deterministic, and yet you cannot give me the number of a ball that is in the urn, then there is an unresolvable contradiction. Any approach that makes these claims makes no sense to me. What happens to each ball can be determined by the puzzle setup. Any ball is either in or out of the urn,

By way of contrast, the other approach, which traces the fate of every individual ball, noting when it enters the urn, and when it leaves, and that no ball is put in the urn two or more times... entirely explains what happens, when it happens and to what.

A ball can only end up in the urn if you put it there and never remove it, and each ball is put in the urn once, and removed once. In particular, if i = 10k + r, where [imath]0\leq r < 10[/imath], then the ith ball is put in the urn on the (k+1)th interaction, and removed on the ith, after which nothing is done to the ball. All balls are numbered so no balls are in the urn.

I agree that its weird. If someone can poke a hole in the reasoning above, I'm all for hearing them out.

Furanku wrote:Obviously at iteration n, we remove ball n, and thus remove the possibility of it being drawn out at midnight. This continues for every iteration, as you've stated. I have difficulty with your interpretation when n approaches infinity, though. Im sure you'll happily correct me when you come back to this, and Im sure that I've misunderstood you or others. To me, thought, it appears that at each iteration n (in question a), the ball n+1 remains in the jar, as do every other ball above it up to 10n. For any arbitrarily large number, this is true (Iteration Googolplex: the googolplexth ball is taken other, but ball 'One googolplex and one' remains, as do all balls between it and 'Ten googolplexes' inclusive), unless I've made an error which you will undoubtedly highlight for me. Now, it looks as if, from your point of view when we cross the mythical 'finite to infinite' iteration boundary, we suddenly get an admittedly silly '[imath]\infty-\infty,[/imath] = 0 balls left' scenario even though, at every finite point, we had a finite number of balls remaining in the jar. This finite number was on the rise too. As soon as you make n = infinity though, finite numbers go out the window, and asking for finite values becomes a bit pointless. The only way we can deal with this is by operating with 'infinities'. Im not even going to try to understand alephs or anything, but it seems to me there is going to be an infinite number of balls remaining in the jar, even if you cannot define the number on any ball picked from the jar after midnight. Whatever undefined quantity of balls that is removed at midnight, 9 times that undefined quantity remains. While trying to multiply infinity by anything is patently silly, so is trying to use remove infinity balls from an infinitly large jar filled with an infinite number of balls.

Dealing with infinity is always a bit ridiculous. And I can't remember what I was saying.

For any arbitrarily large natural Number N it will be taken out of the Jar at Iteration N, which is guaranteed to happen before midnight. If you state that some arbitrarily large numbered Ball is still in the Jar at the end I can tell you exactly when it is removed. There also cannot be any balls with some of those not closer defined infinite numbers in the Jar because we only ever put balls with finite numbers into the Jar (Bolded for emphasis, this is important). At any given Step N we put in the numbers 10*N - 9 to 10*N, and all these numbers are obviously finite. If you claim that some balls with some other labels are in the jar at the end, you have to state at what point they are actually put into it.

The solution to this problem is counterintuitive, that's why it is interesting in the first place.

See, thats where I have the problem. After the nth iteration, where ball n is removed, ball (n+1) still remains.Define for me the final iteration, at midnight, and I'll tell you what the lowest number on a ball in the jar is.

EDIT: Agreeing to classify infinity (an undefined number) as a valid number of iterations (which you guys seem to suggest) leads to a situation where the number upon any ball cannot be defined. Balls exist in the jar, but their numbers are undefined. I could easily ask you to tell me the value of y in y=x+1 as x approaches infinity. Whether you can give an EXACT answer is extremely unlikely, though. Asking of me the exact same thing is pretty suspect.

Each step, you add nine balls (not ten), and instead of removing the lowest numbered ball, you get out a pen and add a '0' to the end of its number. So you add 1-9, and the 1 becomes 10; second step add 11-19, and the 2 becomes 20. So, at every step, the state of the jug (the numbers on the balls it contains) will be identical to case A, but since you never remove any balls, it seems even more nonsensical to claim the jug is empty at midnight. And yet...

Thank you! I think my modifications were broken, but this one is beautiful.

jestingrabbit wrote:I would actually argue that there are an infinite number of balls in the urn at the end of this process, but the "numbers" on them don't have a finite number of digits: they'll have an arbitrary string of digits followed by an infinite string of 0's.

That's just silly... "147", "1470", "147000..." are just symbols. What if we added 5 balls each time and multiplied the lowest by 6? Since multiplying by 6 doesn't have a nice pattern in base 10, you can't tell me what the numbers at the end will look like.

Note that I don't necessarily think that the logic for the original problem is wrong, it's just that you seem to be biting a rather large bullet here: The state of the urn is identical at the end of every step (between the original and Chicken's modification), and yet they end up in different places at infinity? The bullet that you're biting is that the way we arrive at a position can affect what happens in the end. I'm still hideously confused.

EDIT: I can design problems that hurt my brain even more. Like, what if on odd numbered steps you add ten balls and remove the lowest, and on even numbered steps you add nine balls and renumber the lowest in the way Vicious Chicken suggested. I suspect JR will say:The urn has an infinite amount of balls in it, numbered with arbitrary strings ending in an even digit, followed by an infinite string of zeros.

Indigo is a lie.Which idiot decided that websites can't go within 4cm of the edge of the screen?There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

Each step, you add nine balls (not ten), and instead of removing the lowest numbered ball, you get out a pen and add a '0' to the end of its number. So you add 1-9, and the 1 becomes 10; second step add 11-19, and the 2 becomes 20. So, at every step, the state of the jug (the numbers on the balls it contains) will be identical to case A, but since you never remove any balls, it seems even more nonsensical to claim the jug is empty at midnight. And yet...

Thank you! I think my modifications were broken, but this one is beautiful.

jestingrabbit wrote:I would actually argue that there are an infinite number of balls in the urn at the end of this process, but the "numbers" on them don't have a finite number of digits: they'll have an arbitrary string of digits followed by an infinite string of 0's.

That's just silly... "147", "1470", "147000..." are just symbols. What if we added 5 balls each time and multiplied the lowest by 6? Since multiplying by 6 doesn't have a nice pattern in base 10, you can't tell me what the numbers at the end will look like.

Well, then there are an infinite number of balls in the urn, but the digits that are written on them are not well defined (unless something weird happens when you repeatedly multiply by 6 which isn't obvious).

Macbi wrote:Note that I don't necessarily think that the logic for the original problem is wrong, it's just that you seem to be biting a rather large bullet here: The state of the urn is identical at the end of every step (between the original and Chicken's modification), and yet they end up in different places at infinity? The bullet that you're biting is that the way we arrive at a position can affect what happens in the end. I'm still hideously confused.

imo, that is most assuredly the case. How you get there changes everything. Consider the sum 1-1+1-1+1-1+1-1+1... if you bracket it one way, you get 0, another way, you get 1, no brackets and you don't have a converging sequence. How you go to infinity changes the outcome.

Macbi wrote:EDIT: I can design problems that hurt my brain even more. Like, what if on odd numbered steps you add ten balls and remove the lowest, and on even numbered steps you add nine balls and renumber the lowest in the way Vicious Chicken suggested. I suspect JR will say:The urn has an infinite amount of balls in it, numbered with arbitrary strings ending in an even digit, followed by an infinite string of zeros.

To give a closer look at it: While the state of the urn is identical in both scenarios, the individual fate of each ball is immensly different: In the original scenario, each ball is interacted with exactly twice: Once to put it in, and once to take it out again later. I think most of us agree now that the jar is empty at the end because every ball that is put in is taken out again eventually.

Now what happens in the relabeling case? Each ball is put in once, and then is interacted upon infinitly often, putting another trailing zero on it each time. No ball is ever taken out, so the Balls will still be in the Jar at the end. What changes is their label, though. What you do by relabeling the lowest ball is effectively taking out the number it represents, you however do not remove the actual ball. What still holds true in this scenario is that you are removing every natural number eventually, so there won't be any natural numbers on the balls that remain at the end. The balls however are still there, and because you added an infinite number of zeros to the end of each label you get to jestingrabbit's solution. Note how the strings on the balls do not directly represent a natural number anymore because they are nonterminating. So you still have balls left at the end, but no numbers on them.

Now, if a ball is still in the urn after every step, then its in the urn in the the end. But, if there is some step such that after that step the ball is never in the urn, then, the ball can't be in the urn at the end.

Exactly. What's going on here and confusing people is that[math]|\lim_{n \to \infty} A_n|=0,[/math]but[math]\lim_{n \to \infty} |A_n|=\infty.[/math]This is not the least bit paradoxical, but is a simple conclusion resulting from the definition of limit, and definition of the sets [imath]A_n[/imath]. If you accept that a ball is in the jug at the end if and only if at some point it is put in and not later removed, you are forced into the conclusion that the correct answer is given by [imath]|\lim_{n \to \infty} A_n|=0[/imath] rather than [imath]\lim_{n \to \infty} |A_n|[/imath]. As for blacksail's objection that "at every step you add balls and the sum of positive numbers is positive," I merely point out that there's no reason why the number of balls in the jug at the end should be the limit of the numbers in the ball at the jug at stages along the way. All the balls that enter the jug eventually leave it. It's true that balls are entering faster than they are leaving, but since every ball that enters the jug also leaves it, the jug is empty.

Vicious Chicken wrote:A slight variation to case A:

Each step, you add nine balls (not ten), and instead of removing the lowest numbered ball, you get out a pen and add a '0' to the end of its number. So you add 1-9, and the 1 becomes 10; second step add 11-19, and the 2 becomes 20. So, at every step, the state of the jug (the numbers on the balls it contains) will be identical to case A, but since you never remove any balls, it seems even more nonsensical to claim the jug is empty at midnight. And yet...

I would actually argue that there are an infinite number of balls in the urn at the end of this process, but the "numbers" on them don't have a finite number of digits: they'll have an arbitrary string of digits followed by an infinite string of 0's.

That was also the solution I came up with. As for Macbi's objection:

Macbi wrote:"147", "1470", "147000..." are just symbols. What if we added 5 balls each time and multiplied the lowest by 6? Since multiplying by 6 doesn't have a nice pattern in base 10, you can't tell me what the numbers at the end will look like.

It was explicitly stated in the problem that we took out a pen and added a '0' on the end, so treating them as just symbols is entirely appropriate. Also, the state of the urn at the ends of each stage is not the same as in situation a) above, because the same ball is numbered 10 at the end of stage 1, 100 at the end of stage 2, 1000 at the end of stage 100, and so on - it just has different ink on it at those stages. If you "added 5 balls and multiplied the lowest by six", then it depends what you mean by "multiplied the lowest by 6". If you cross out the old number and write a new number which is six times the old number, then the (infinitely many) balls left at the end would all have an infinite sequence of crossed out numbers. If you always erased the old number and wrote the new one in its places, you would have infinitely many blank balls, applying the same logic to the ink: all ink added to each ball is later removed, so in the end the balls are blank. (Assuming perfect erasers. With real-life erasers, you would have a really ugly smudge.)

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

and the limit equals whatever set those sets equal. Basically, when skep insists that "a ball is in the jug at the end if and only if at some point it is put in and not later removed", he is using the LHS of that equation.

skeptical scientist wrote:If you always erased the old number and wrote the new one in its places, you would have infinitely many blank balls, applying the same logic to the ink: all ink added to each ball is later removed, so in the end the balls are blank. (Assuming perfect erasers. With real-life erasers, you would have a really ugly smudge.)

I don't think that's quite right, I think we're looking at things through different topologies though, and either is probably valid to some extent... hmmm... time to sleep.

Edit: no, its not topologies. When you erase and rewrite you're doing something within a single time step. When at one step you put a ball in, and then remove it at a later step, you're doing something over a collection of steps. The sequence that we're taking the limit of consists of what is left after each time step, not what goes on during the time step.

jestingrabbit wrote:When you erase and rewrite you're doing something within a single time step. When at one step you put a ball in, and then remove it at a later step, you're doing something over a collection of steps. The sequence that we're taking the limit of consists of what is left after each time step, not what goes on during the time step.

But it's different ink, just as its different balls. If it were the same ink, only moved, we'd have a problem of the limit not necessarily existing, but since it's different ink we can use the same argument as with the balls. In any case, artificially designating certain moments as "ends of steps" and only considering what happens at those moments is highly suspect.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

Let's take the Ink confusion one step further. If you completely erase the label and then write on the new one afterwards it at least seems plausible that in the limit, the balls end up in the state of being unlabeled, as they pass through it an infinite number of times (and so is a limit point of the label sequence). How about instead first erasing the first half of the label, then writing down the new label in its place, then erasing the remaining half of the original label. Skeptical Scientist's argument should still hold for this variation, but no ball is ever in a state where it is entirely unlabeled, yet it seems to not bear a label at the end. I can't offer an explanation for this, and it seems more paradoxical than any of the other issues we came accross so far.

If you consider the last infinitesimal bit of time, rather than a number of time steps, you add and take out an infinite number of balls. Inf-inf is not well defined, which is why we have multiple answers.

LLCoolDave wrote:Let's take the Ink confusion one step further. If you completely erase the label and then write on the new one afterwards it at least seems plausible that in the limit, the balls end up in the state of being unlabeled, as they pass through it an infinite number of times (and so is a limit point of the label sequence). How about instead first erasing the first half of the label, then writing down the new label in its place, then erasing the remaining half of the original label. Skeptical Scientist's argument should still hold for this variation, but no ball is ever in a state where it is entirely unlabeled, yet it seems to not bear a label at the end. I can't offer an explanation for this, and it seems more paradoxical than any of the other issues we came accross so far.

It's no more or less paradoxical than the balls in the jug; the jug is never empty at any stage along the way, and yet ends up empty (under one interpretation anyways, the only one I think is valid). The same logic applies, so long as you think of each new bit of ink as different, just as you think of each new ball as different. First molecules 1 through 6e23 of ink are added. Then they are removed and molecules 6e23+1 through 1.2e24 are added. Eventually every molecule of ink which is added is removed, so the situation is exactly analogous with the balls.

BlackSails wrote:Im changing my answer from "full of balls" to "DNE"

If you consider the last infinitesimal bit of time, rather than a number of time steps, you add and take out an infinite number of balls. Inf-inf is not well defined, which is why we have multiple answers.

It's true that [imath]\infty - \infty[/imath] is not well defined, but that doesn't mean you can't attach sizes to sets which are the difference of two infinite sets. The composite numbers are the natural numbers minus the prime numbers, but there's no problem saying how many composite numbers there are. Similarly, the balls left in the jug are the balls that are placed in the jug minus the balls that are removed from the jug, and it shouldn't be hard to see what this set is.

Certainly infinity, infinite sequences, and limits can all be confusing, and there are situations not too different from this which are impossible to analyze. But here, I would argue, the situation is really quite simple:

There are balls, labeled 1, 2, 3, and so on, and no other balls.

Each of the balls 1, 2, 3, and so on is first put in the jug and then later removed from the jug, and is never heard from again.

Therefore, the jug does not contain ball 1, ball 2, or any of the other numbered balls. Since those are the only balls, the jug winds up empty.

The fact that the jug is never empty along the way - that it, in fact, gets more and more full as midnight approaches - is completely irrelevant.

p.s. I'm giving up spoilering the discussion of puzzle a), since it just gets in the way, and everyone who has read this far is probably not bothered by it. If anyone has a problem with this, let me know, and I'll go back and edit it.

I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

I think I've managed to dissolve the paradox (at least, I no longer feel confused by it, which is a good sign). I believe the problem is that we currently have no mathematical way of formally talking about time. Once you add one the problem becomes boring.

Okay then, I define time to be a function f(t) from the Reals to the possibility-space of the urn-ball system. For example:

f(-10)="Balls from one to ten are in the urn"f(-5)="Balls from two to twenty are in the urn"f(-4)="Balls from two to twenty are in the urn"f(-2.5)="Balls from three to thirty are in the urn"

And the problem asks us to find f(0)

In the problem statement we are told f(t) for various t's, and we know that these are the only times that the contents of the urn change. This means that if we want to find out what the contents of the urn was at some time where it was not specified, we simply look back to see when it was last specified and make the assumption that it hasn't changed. For example, if we want to find f(-1) we look back for the last point t was -10 times a power of two (i.e. -1.25) and take f(t) at that point: "Balls from four to forty are in the urn". The problem comes when we look back an find that there is no last point at which a change was made. This can happen in the boring case where we, say, want to find f(-11), in which case we can fudge it by assuming that f(-[imath]\infty[/imath])="Urn is empty". The more exiting case comes when we want f(0) and we find that because an infinite amount of changes occurred before t=0, there was no last one to occur. With the tools we have so far we can't find f(0), we need some new "Supertask Axiom" that will tell us what to do. We can basically formulate this axiom to be whatever we like, but it would be better if it had some nice properties.

However, we can see from other supertask paradoxes such as Thompson's Lamp that no such nice axiom exists. To reformulate it in terms of balls and urns:When t=-22n for any integer n, f(t)="Ball one is in the urn", when t=-22n+1 for any integer n, f(t)="Ball two is in the urn". Because of the symmetry in this example no Supertask Axiom is going to provide an adequate answer for f(t) when t=0. (although I do like to imagine that JR will claim that there is one ball with "1.5" written on it )

So having seen that we need a Supertask Axiom, but that none are adequate, we can safely call the problem stupid. So why does it seem like the problem (i.e. the original problem a)) has a sensible answer? Well it turns out that although the entire urn-ball system performs a supertask, each individual ball does not. Note that this does not happen in my Thompson's Lamp formulation. This means that we can solve some types of urn-ball paradox with the Partial Supertask Axiom:"If the possibility-space of a system performing a supertask factorises into smaller possibility-spaces of systems that do not perform supertasks, then the position of the large system is the union of the positions of the smaller systems"

This is what JR and SkepSci are going on about when they say "Just consider each individual ball"

Indigo is a lie.Which idiot decided that websites can't go within 4cm of the edge of the screen?There should be a null word, for the question "Is anybody there?" and to see if microphones are on.

Goplat wrote:Let's turn this 169-ish problem into something that actually makes mathematical sense. Let p(i,n) = probability of ball i being in the jug after n rounds. The "correct" solutions to this problem are really solutions to [math]\sum_{i=1}^{\infty}\lim_{n\rightarrow\infty}p(i,n)[/math] while some people instead interpreted it as [math]\lim_{n\rightarrow\infty}\sum_{i=1}^{\infty}p(i,n)[/math] What makes one interpretation better than the other?

Secondly, we shouldn't worry about the probabilistic problem c. Blacksails and Furanku have issues with the solution that bogglesteinsky presented for problem a, and there are no probabilities there, so why don't we discuss the disagreement in the simplest situation in which the disagreement occurs.

For deterministic problems a and b, obviously p(i,n) is always 0 or 1. I defined the function in terms of probability because that can work for all three problems.

I would just point out, though, that just because there is a calculation that can be done, doesn't mean that the calculation has meaning (the case of the missing dollar has a good example of a calculation that has no meaning).

In this case, neither calculation has meaning; the problem corresponds to nothing that could happen in real life. If it specified where the limit was supposed to go, we could solve it anyway, but it doesn't, making it ambiguous.

If every ball has a number on it, and you claim that there are an infinite number of balls in the urn, and what we are doing is deterministic, and yet you cannot give me the number of a ball that is in the urn, then there is an unresolvable contradiction. Any approach that makes these claims makes no sense to me.

It goes both ways. If you claim that there are zero balls in the urn as a result of repeatedly adding 9 balls, that makes no sense to me.

By way of contrast, the other approach, which traces the fate of every individual ball, noting when it enters the urn, and when it leaves, and that no ball is put in the urn two or more times... entirely explains what happens, when it happens and to what.

It certainly doesn't explain how the infinite series 9+9+9+9+... converges and equals 0.

The problem is too poorly defined to be solvable, because the wording doesn't eliminate the "wrong" method of solving it.

Goplat wrote:If you claim that there are zero balls in the urn as a result of repeatedly adding 9 balls, that makes no sense to me.

...except you're not repeatedly adding 9 balls. You are repeatedly adding and removing balls, so that every ball which is added is later removed. Why should it matter that the number of balls in the urn is increasing during the procedure, if every ball which is added is later removed?

Macbi wrote:Well it turns out that although the entire urn-ball system performs a supertask, each individual ball does not. This means that we can solve some types of urn-ball paradox with the Partial Supertask Axiom.

Exactly. There are other similar situations which are impossible to analyze, and this shouldn't worry us. But our situation can be understood using a ball-by-ball analysis, or what you call the "partial supertask axiom".